Solution:

Let $t = \log_{2026} x.$ Taking $\log_{2026}$ of both sides of $x^{(\log_{2026} x)/20} = 26x$ gives $$\frac{t^2}{20} = \log_{2026} 26 + t, \qquad \text{that is} \qquad t^2 - 20t - 20\log_{2026} 26 = 0.$$ The discriminant $400 + 80\log_{2026} 26$ is positive, so there are two real roots $t_1, t_2,$ each giving a valid positive solution $x = 2026^{t}.$

By Vieta's formulas $t_1 + t_2 = 20,$ so the product of the solutions is $2026^{t_1} \cdot 2026^{t_2} = 2026^{20}.$ Since $2026 = 2 \cdot 1013$ and $1013$ is prime, $P = 2^{20} \cdot 1013^{20}$ has $21 \cdot 21 = 441$ positive divisors.